I’m given this problem:
Exercise. Suppose $f$ is analytic in the whole complex plane, except for an isolated singularity of unspecified kind at $2+2i$, and suppose $\gamma$ is the circumference of radius $2$ centered at $3$. Expressing the result in terms of values of $f$ or of its derivatives, calculate the following integrals (when possible): $$\int_\gamma \frac{f(z)\sqrt z}{(z+i)^2} dz, \quad \int_\gamma \frac{f(z)\sqrt z}{(z-1)^2} dz, \quad \int_\gamma \frac{f(z)\sqrt z}{(z-4)^2} dz. $$
The contour does not enclose the singularity of $f$, nor any singularity of $\sqrt z$ (taking the branch cut at the negative real axis), so $g(z)=f(z)\sqrt z$ is analytic in the interior of $\gamma$. I think I am to apply the Cauchy integral formula to $g$, $$g’(z_0) = \frac 1 {2\pi i}\int_\gamma \frac{g(z)}{(z-z_0)^2}dz =: I(z_0),$$ choosing $z_0 = -i,1,4$. For the last case, I’ve found $$I(4) = 2\pi i g’(4) = 4\pi i f’(4) - \frac 1 2\pi i f(4)$$ However, for the other cases, we have either $z_0$ on the contour or outside of it, so technically Cauchy’s formula wouldn’t apply. However, $\gamma$ is homotopic to a circumference that encloses all $-i$ and $1$ and $4$, and within which $g$ is analytic: does that mean I’m good to go? That is, are all three integrals computable with the given data, despite the warning at the end the exercise?